How do you find the 6 trigonometric functions for 150 degrees?

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Answer 1 · 2015-09-21T21:21:22

Refer to explanation

We have that

$sin(150)=sin(180-30)=sin30=1/2$

$csc(150)=1/sin(150)=2$

$cos (150) = –cos(30) =-sqrt3/2$

$sec(150) = 1/cos(150)=-2/sqrt3$

$tan(150)=-tan(30)=-sqrt3/3$

$cot(150)=1/tan(150)=-sqrt3$